Evening from WA,
Design example for the conventional fighter:
"Design Example: p. 191
" Robert’s ’Mechbuster—as a conventional fighter weighing 50 tons—has a maximum armor limit of 50 points (50 tons x 1 = 50 points). Choosing standard armor (to keep costs low), Rob finds that the weight for maximum armor protection comes to 3.5 tons (50 points ÷ 16 = 3.125, rounded up to 3.5). Deciding to place the maximum amount of armor on his fighter, Rob allocates 18 points to the nose, 11 points to each wing and 10 points to the aft, for a total of 50 points allocated (18 + 11 + 11 + 10 = 50). With 3.5 tons invested in its armor, the ’Mechbuster now has 14.5 tons remaining (1s.5 = 14.5)."
While reviewing Aerospace Unit Construction Step 3: Add Armor the 50 ton MechBuster Conventional Fighter's 3.5 tons of Standard armor exceeds the Maximum Armor Levels Table requirement on p. 191.
1. From Step 3: Add Armor p. 190: Second paragraph.
"The maximum amount of armor an aerospace unit may mount is based on the unit’s type and body shape (conventional fighter, aerospace fighter aerodyne Small Craft/DropShip or spheroid Small Craft/DropShip) as well as its weight and structural integrity. This value—applicable for Clan and Inner Sphere aerospace units—is shown in the Aerospace Unit Maximum Armor Levels Table below (p. 191). Regardless of what kind of armor the unit uses, aerospace units "may not exceed their maximum point value" (or armor tonnage for larger units). Note, however, that this value reflects the aerospace unit’s maximum total armor; the number of points assigned per location is left up to the designer’s discretion."
From the Aerospace Unit Maximum Armor Levels Table p. 191
Unit Type: Conventional Fighter
Maximum Armor Level: (Tonnage x 1)points
Armor Facings: Nose, Left Wing, Right Wing, Aft
The MechBuster is a 50 ton conventional fighter and the Maximum Armor Level = (50 x 1) points = 50 Armor Points.
2. First full paragraph first sentence on p. 191:
"With the Points per Ton value determined, finding the amount of points provided for a given tonnage is a matter of multiplying that value by the desired tonnage of armor and then rounding the final result down to the nearest whole number."
Multiplying 3.5 tons x 16 = 56 Armor Points which exceeds the fighter's maximum armor level.
3. From p. 191 "If the designer wants a specific number of armor points, he may simply divide that number of points by the armor type’s Points per Ton value and round the result up to the nearest half-ton. For example, if an aerospace unit is to mount 5 tons of standard armor, the designer can multiply 5 by 16 to find the result (80 points). If, on the other hand, the designer wants 170 points of standard Inner Sphere ferro-aluminum armor, he should divide the armor value of 170 by 17.92 (Points per Ton value for standard Inner Sphere ferro-aluminum) to find the tonnage needed (9.5 tons; 170 ÷ 17.92 = 9.487, rounded up to 9.5)."
The conventional fighter example used the method described above: "Choosing standard armor (to keep costs low), Rob finds that the weight for maximum armor protection comes to 3.5 tons (50 points ÷ 16 = 3.125, rounded up to 3.5)."
I have a copy of the HeavyMetal application Aero v 1.06 R7 and I pulled up the file for a MechBuster 50 ton Conventional Fighter the deign window lists Maximum Armor Points level as 50 with 3 tons of armor providing 48 Armor Points.
The described rounding up to the nearest have in the MechBuster's case exceeds the Maximum Protection Level rule.
I think that in the case of the MechBuster the calculated 3.125 tons of weight should be rounded down to 3 tons versus 3.5 tons.
I have checked the TM errata and did not see an entry should I suggest a correction in the Errata page?
